ctg^2x>1 inequation

Given the inequality:
$$\cot^{2}{\left(x \right)} > 1$$
To solve this inequality, we must first solve the corresponding equation:
$$\cot^{2}{\left(x \right)} = 1$$
Solve:
Given the equation
$$\cot^{2}{\left(x \right)} = 1$$
transform
$$\cot^{2}{\left(x \right)} - 1 = 0$$
$$\cot^{2}{\left(x \right)} - 1 = 0$$
Do replacement
$$w = \cot{\left(x \right)}$$
This equation is of the form

a*w^2 + b*w + c = 0

A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$w_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$w_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 1$$
$$b = 0$$
$$c = -1$$
, then

D = b^2 - 4 * a * c = 

(0)^2 - 4 * (1) * (-1) = 4

Because D > 0, then the equation has two roots.

w1 = (-b + sqrt(D)) / (2*a)

w2 = (-b - sqrt(D)) / (2*a)

or
$$w_{1} = 1$$
$$w_{2} = -1$$
do backward replacement
$$\cot{\left(x \right)} = w$$
substitute w:
$$x_{1} = - \frac{\pi}{4}$$
$$x_{2} = \frac{\pi}{4}$$
$$x_{1} = - \frac{\pi}{4}$$
$$x_{2} = \frac{\pi}{4}$$
This roots
$$x_{1} = - \frac{\pi}{4}$$
$$x_{2} = \frac{\pi}{4}$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} < x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{\pi}{4} - \frac{1}{10}$$
=
$$- \frac{\pi}{4} - \frac{1}{10}$$
substitute to the expression
$$\cot^{2}{\left(x \right)} > 1$$
$$\cot^{2}{\left(- \frac{\pi}{4} - \frac{1}{10} \right)} > 1$$

   2/1    pi\    
cot |-- + --| > 1
    \10   4 /    

Then
$$x < - \frac{\pi}{4}$$
no execute
one of the solutions of our inequality is:
$$x > - \frac{\pi}{4} \wedge x < \frac{\pi}{4}$$

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        /     \  
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       x1      x2